HDU 5437 Alisha’s Party (优先队列)——2015 ACM/ICPC Asia Regional Changchun Online
Each time when Alisha opens the door, she can decide to let p people enter her castle. If there are less than p people in the lobby, then all of them would enter. And after all of her friends has arrived, Alisha will open the door again and this time every friend who has not entered yet would enter.
If there are two friends who bring gifts of the same value, then the one who comes first should enter first. Given a query n Please tell Alisha who the n−th person to enter her castle is.
In each test case, the first line contains three numbers k,m and q separated by blanks. k is the number of her friends invited where 1≤k≤150,000. The door would open m times before all Alisha’s friends arrive where 0≤m≤k. Alisha will have q queries where 1≤q≤100.
The i−th of the following k lines gives a string Bi, which consists of no more than 200 English characters, and an integer vi, 1≤vi≤108, separated by a blank.Bi is the name of the i−th person coming to Alisha’s party and Bi brings a gift of value vi.
Each of the following m lines contains two integers t(1≤t≤k) and p(0≤p≤k) separated by a blank. The door will open right after the t−th person arrives, and Alisha will let p friends enter her castle.
The last line of each test case will contain q numbers n1,...,nq separated by a space, which means Alisha wants to know who are the n1−th,...,nq−th friends to enter her castle.
Note: there will be at most two test cases containing n>10000.
考点明确,就是优先队列,再模拟一下可以出结果。每入队一个人就操作一次。数据量略大,用c的输入输出过了,cin,cout会TLE。
#include<stdio.h>
#include<iostream>
#include<string>
#include<cstring>
#include<algorithm>
#include<queue> using namespace std; struct fri
{
char name[];
int come,v;
bool operator < (const fri& f) const
{
if(v!=f.v) return v<f.v;
else return come>f.come;
} //为了优先队列里的顺序而重载
}f[]; int c[],ans[]; int main()
{
int k,m,q,t,p,sum;
int T; cin>>T; while(T--)
{
memset(c,,sizeof(c));
sum=;
cin>>k>>m>>q;
for(int i=;i<=k;++i)
{
scanf("%s %d",f[i].name,&f[i].v);
f[i].come=i;
}
for(int i=;i<=m;++i)
{
scanf("%d %d",&t,&p);
c[t]=p;
} //输入顺序不确定,t个人时进行操作 priority_queue<fri> que;
for(int i=;i<=k;++i) //每来一个人,检查是否开门
{
que.push(f[i]); //来第i个人入队
int n = c[i]; //来了i个人时应出队的n个人进行操作
while(n&&!que.empty())
{
ans[sum++]=que.top().come; //记录出队人的序号,下面用于输出名字
que.pop();
--n;
}
}
while(!que.empty()) //剩下的人依次出队
{
ans[sum++]=que.top().come;
que.pop();
} while(q--)
{
int a;
scanf("%d",&a);
printf("%s",f[ans[a]].name);
if(q==) printf("\n");
else printf(" ");
}
}
return ;
}
HDU 5437 Alisha’s Party (优先队列)——2015 ACM/ICPC Asia Regional Changchun Online的更多相关文章
- (并查集)Travel -- hdu -- 5441(2015 ACM/ICPC Asia Regional Changchun Online )
http://acm.hdu.edu.cn/showproblem.php?pid=5441 Travel Time Limit: 1500/1000 MS (Java/Others) Memo ...
- (二叉树)Elven Postman -- HDU -- 54444(2015 ACM/ICPC Asia Regional Changchun Online)
http://acm.hdu.edu.cn/showproblem.php?pid=5444 Elven Postman Time Limit: 1500/1000 MS (Java/Others) ...
- 2015 ACM/ICPC Asia Regional Changchun Online HDU 5444 Elven Postman【二叉排序树的建树和遍历查找】
Elven Postman Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)T ...
- hdu 5444 Elven Postman(二叉树)——2015 ACM/ICPC Asia Regional Changchun Online
Problem Description Elves are very peculiar creatures. As we all know, they can live for a very long ...
- (线段树 区间查询)The Water Problem -- hdu -- 5443 (2015 ACM/ICPC Asia Regional Changchun Online)
链接: http://acm.hdu.edu.cn/showproblem.php?pid=5443 The Water Problem Time Limit: 1500/1000 MS (Java/ ...
- 2015 ACM/ICPC Asia Regional Changchun Online HDU - 5441 (离线+并查集)
题目:http://acm.hdu.edu.cn/showproblem.php?pid=5441 题意:给你n,m,k,代表n个城市,m条边,k次查询,每次查询输入一个x,然后让你一个城市对(u,v ...
- Hdu 5442 Favorite Donut (2015 ACM/ICPC Asia Regional Changchun Online 最大最小表示法 + KMP)
题目链接: Hdu 5442 Favorite Donut 题目描述: 给出一个文本串,找出顺时针或者逆时针循环旋转后,字典序最大的那个字符串,字典序最大的字符串如果有多个,就输出下标最小的那个,如果 ...
- Hdu 5446 Unknown Treasure (2015 ACM/ICPC Asia Regional Changchun Online Lucas定理 + 中国剩余定理)
题目链接: Hdu 5446 Unknown Treasure 题目描述: 就是有n个苹果,要选出来m个,问有多少种选法?还有k个素数,p1,p2,p3,...pk,结果对lcm(p1,p2,p3.. ...
- hdu 5444 Elven Postman(根据先序遍历和中序遍历求后序遍历)2015 ACM/ICPC Asia Regional Changchun Online
很坑的一道题,读了半天才读懂题,手忙脚乱的写完(套上模板+修改模板),然后RE到死…… 题意: 题面上告诉了我们这是一棵二叉树,然后告诉了我们它的先序遍历,然后,没了……没了! 反复读题,终于在偶然间 ...
随机推荐
- System.Runtime.InteropServices.COMException (0x800706BA) 解决方法
提示“操作失败:无法获取MAC地址.”错误的解决方法. .NET 获取 MAC地址可能会遇到 System.Runtime.InteropServices.COMException (0x8007 ...
- 【转】工科男IT职场求生法则
转自:http://www.36dsj.com/archives/3459 我在IT职场打滚超过10年了,从小小的程序员做到常务副总.相对于其它行业,IT职场应该算比较光明的了,但也陷阱重重,本文说说 ...
- 【转】浅析linux内存模型
转自:http://pengpeng.iteye.com/blog/875521 0. 内存基本知识 我们通常称 linux的内存子系统为:虚拟内存子系统(virtual memory system) ...
- 巧妙使用Jquery 改变元素的 onclick 事件
需要点击图片将套组发布, 页面代码: <img width="20px" src=" <s:property value="IMAGES_PATH& ...
- Codeforces Round #338 (Div. 2) C. Running Track dp
C. Running Track 题目连接: http://www.codeforces.com/contest/615/problem/C Description A boy named Ayrat ...
- 如何利用PhoneGap制作地图APP
摘要:百度地图API是一套由javascript编写的地图程序接口,按说它应该运行在浏览器上.现在,只要利用PhoneGap,我们就能开发出移动平台上能使用的APP了! --------------- ...
- php调试小技巧
/** * 用来调试输出结果 * @param type $data * @return type */ function shionyu_debug($data) { ob_start(); var ...
- 比较全面的MySQL优化参考
本文整理了一些MySQL的通用优化方法,做个简单的总结分享,旨在帮助那些没有专职MySQL DBA的企业做好基本的优化工作,至于具体的SQL优化,大部分通过加适当的索引即可达到效果,更复杂的就需要 ...
- cocos2d-x在android下的编译
$(call import-add-path,E:/cocos2d-2.0-x-2.0.3) include $(BUILD_SHARED_LIBRARY) http://www.cnblogs.co ...
- 如何在HTML5 图片预览
HTML5的 File API允许浏览器访问本地文件系统,借助它我们可以实现以前无法实现的本地图片预览功能. 先介绍下该API实现了那些接口: 1.Blob接口,表示原始的二进制数据,通过它可以访问到 ...